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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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Rational box-step functions form a countable dense subset of Lp(Rn) for 1p<

Statement

Assume the Axiom of Countable Choice.

Let 1p<. The finite linear combinations of indicator functions of half-open boxes with rational endpoints and rational coefficients form a countable dense subset of Lp(Rn).

Facts & Assumptions

Given: The Axiom of Countable Choice, 1p< and ε>0.

[L5]

A space is separable exactly when it has a countable dense subset (Separability: the existence of an at most countable dense subset).

Proof

technique · direct
1.1

Let R be the family of half-open boxes [L2, L4, given, algebra] i=1n(ai,bi] with rational endpoints. By [L2] there are countably many such boxes, and by [L4] the set S of all finite rational linear combinations of indicators 1R with RR is countable.

L2L4givenalgebra
2.1

By [L1], it is enough to approximate a single box indicator. So fix a [L1, L3, step 1.1, choose, algebra] δ>0 and let B=i=1n(αi,βi]. Choose rationals ai<αi<βi<bi so close to the endpoints that, with M:=1+maxi{ai,bi,αi,βi}, 2n(2M)n1maxi{(αiai),(biβi)}<δ. Then BR is contained in the union of the 2n coordinate slabs where one coordinate lies in (ai,αi] or (βi,bi] while the others stay in [M,M]. By [L3], each slab has measure at most (2M)n1maxi{(αiai),(biβi)}, so λn(BR)<δ for the rational box R:=i(ai,bi]. Hence 1B1Rp=λn(BR)1/p<δ1/p, which can be made arbitrarily small; approximating finitely many coefficients by rationals then makes every box-step function arbitrarily close to an element of S.

L1L3step 1.1choosealgebra
3.1

Therefore S is countable and dense. By [L5], [L5, step 1.1, step 2.1] Lp(Rn) is separable, with S as an explicit dense subset.

L5step 1.1step 2.1

Depends on

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